Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 20G / Solution

Use the final printed restriction that $m$ is even. Square-freeness then gives $m\equiv 2(\mathrm{mod}4)$, and the integer ring is ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{-m}]$. An integer solution has $x>0$. If $y$ were even, $y^2+m$ would have 2-adic valuation one, impossible for a cube; hence $x,y$ are odd. For an odd prime $p\mid m$, $p\mid y$ would similarly give valuation one. Thus $\gcd (y,m)=1$.

The ideals generated by $y+\sqrt{-m}$ and $y-\sqrt{-m}$ are coprime. Any common prime ideal divides their sum and difference and therefore lies above a rational prime dividing $2m$ and $y$. The odd possibilities have just been excluded; above two, $\sqrt{-m}$ is zero modulo the ramified prime while $y$ is odd, so neither factor is divisible by it. Unique factorization of ideals in a number field innow yields $(y+\sqrt{-m})={\mathfrak{a}}^3$. Since the class number is coprime to three, the ideal class of $\mathfrak{a}$ is trivial. The ideal is principal, and the only units for this imaginary quadratic field with $m\ge 5$ are $\pm 1$, which can be absorbed into a cube. HenceComparing coefficients of $\sqrt{-m}$ givesThus $v=1$ with $3u^2=m+1$, or $v=-1$ with $3u^2=m-1$. Both alternatives cannot occur, since $m-1$ and $m+1$ cannot both be divisible by three. The eligible alternative permits at most the two choices $u=\pm \sqrt{(m\pm 1)/3}$, and thenConsequently there are at most two integer solutions $(x,y)$, which, when present, have the same $x$ and opposite $y$. The proof uses the evenness assumption; it does not silently claim the same coprimality argument for every odd $m$.